= Solution
\b[Classical translational statistics.] Write the <number density> as $n=N/V$ and the <thermal de Broglie wavelength> as $\lambda=\sqrt{2\pi\hbar^2/(mk_BT)}$. Quantum exchange is negligible when <wave packets> rarely overlap:
$$
\boxed{n\lambda^3\ll1\quad\text{or}\quad n^{-1/3}\gg\lambda.}
$$
The equivalent state-counting statement is that the mean occupation of each thermally relevant one-particle state is much smaller than one. The equations in this question count one internal species; with $g$ equally populated <spin> states the relevant density per species is $n/g$.
Let $z=e^{\beta\mu}$ be the dimensionless <quantum gas fugacity>, with the lowest one-particle <energy> chosen as zero. Both quantum distributions can be written
$$
n_\eta(E)=\frac1{z^{-1}e^{\beta E}-\eta}
=\frac{ze^{-\beta E}}{1-\eta ze^{-\beta E}},
\qquad\eta=\begin{cases}+1&\text{bosons},\\-1&\text{fermions}.\end{cases}
$$
If $z\ll1$, then $n_\eta(E)\simeq ze^{-\beta E}\ll1$ for all $E\geq0$, recovering <Maxwell–Boltzmann statistics>. Conversely, low occupation even at the spectrum's bottom requires $z/(1-z)\ll1$ for <bosons> or $z/(1+z)\ll1$ for <fermions>; either condition implies
$$
\boxed{z=e^{\beta\mu}\ll1.}
$$
A single high-energy state can be sparsely occupied even in a degenerate gas, so the condition must hold across the relevant spectrum, not just for one arbitrarily chosen <energy>. Also, classical translational statistics does not imply that all internal excitation <energies> or other quantum effects can be ignored.
\b[Grand <partition function>, density and <pressure>.] Noninteracting particles have independent occupations of each one-particle state. Put $x_{\mathbf k}=ze^{-\beta\varepsilon_{\mathbf k}}$. Summing the allowed occupations gives
$$
\Xi_{\mathbf k}=\begin{cases}
\sum_{r=0}^\infty x_{\mathbf k}^r=(1-x_{\mathbf k})^{-1}&\text{bosons},\\
1+x_{\mathbf k}&\text{fermions},
\end{cases}
\qquad
\Xi=\prod_{\mathbf k}(1-\eta x_{\mathbf k})^{-\eta}.
$$
Therefore
$$
\ln\Xi=-\eta\sum_{\mathbf k}\ln(1-\eta ze^{-\beta\varepsilon_{\mathbf k}}).
$$
The <grand potential> is $\Omega_G=-k_BT\ln\Xi=-pV$ for this homogeneous gas. Replacing the sum by $V\int d^3k/(2\pi)^3$ and using $\varepsilon_{\mathbf k}=\hbar^2k^2/(2m)$ gives
$$
\boxed{\beta p=-\eta\int\frac{d^3k}{(2\pi)^3}
\ln\left[1-\eta z e^{-\beta\hbar^2k^2/(2m)}\right].}
$$
The mean number is $N=z\,\partial_z\ln\Xi$, giving
$$
\boxed{\frac NV=\int\frac{d^3k}{(2\pi)^3}
\frac1{z^{-1}e^{\beta\hbar^2k^2/(2m)}-\eta}.}
$$
For <bosons>, the grand sum requires $z<1$ when the ground <energy> is zero. A macroscopic condensate at the limiting $z\to1$ must be separated from the continuum <integral> in the condensed phase. The following small-$z$ calculation is in the normal, nondegenerate regime, so that complication does not enter.
\b[Density expansion and inversion.] Expand the occupation as a <geometric series>:
$$
n_\eta(E)=ze^{-\beta E}+\eta z^2e^{-2\beta E}+z^3e^{-3\beta E}+\cdots.
$$
The Gaussian <wavevector> <integral> for its $\ell$th term is
$$
\int\frac{d^3k}{(2\pi)^3}e^{-\ell\beta\hbar^2k^2/(2m)}
=\left(\frac{m}{2\pi\ell\beta\hbar^2}\right)^{3/2}
=\frac1{\lambda^3\ell^{3/2}}.
$$
It follows that
$$
\boxed{n\lambda^3=z+\frac{\eta z^2}{2^{3/2}}+O(z^3).}
$$
For the inverse series, put $y=n\lambda^3$ and try $z=y+c y^2+O(y^3)$. Substitution gives $0=(c+\eta/2^{3/2})y^2+O(y^3)$, hence
$$
\boxed{z=n\lambda^3-\frac\eta{2^{3/2}}(n\lambda^3)^2+O((n\lambda^3)^3).}
$$
The leading equality $n\lambda^3\simeq z$ directly connects the density and occupancy criteria for classical behavior.
\b[<Pressure> and its first quantum correction.] Expanding the <logarithm> in the <pressure> expression gives one extra factor $1/\ell$ compared with the density series:
$$
\beta p\lambda^3=z+\frac{\eta z^2}{2^{5/2}}+O(z^3).
$$
Insert the inverse series. To second order in $y=n\lambda^3$,
$$
\beta p\lambda^3=y-\frac{\eta y^2}{2^{3/2}}+
\frac{\eta y^2}{2^{5/2}}+O(y^3)
=y-\frac{\eta y^2}{2^{5/2}}+O(y^3).
$$
Thus
$$
\boxed{pV=Nk_BT\left[1-\frac\eta{2^{5/2}}\frac{N\lambda^3}{V}
+O\left((N\lambda^3/V)^2\right)\right].}
$$
The first quantum correction at fixed density and <temperature> is
$$
\boxed{\Delta p=-\frac{\eta}{2^{5/2}}k_BT\,n^2\lambda^3.}
$$
It is negative for <bosons> and positive for <fermions>, and vanishes in the classical density <limit>.
\b[Virial coefficient.] Define the number-density <virial expansion> by
$$
\frac p{nk_BT}=1+B_2(T)n+B_3(T)n^2+\cdots
=\sum_{j=1}^\infty B_j(T)n^{j-1},\qquad B_1=1.
$$
The coefficient $B_j$ has units $\mathrm{volume}^{j-1}$ in this convention. Comparing terms gives
$$
\boxed{B_2^{\mathrm{ex}}(T)=-\frac\eta{2^{5/2}}\lambda^3
=-\frac\eta{2^{5/2}}\left(\frac{2\pi\hbar^2}{mk_BT}\right)^{3/2}.}
$$
This is the <exchange contribution to the second virial coefficient>. It arises without a pair-interaction potential; the underlying gas is noninteracting. A molar-density convention would multiply this $B_2$ by the Avogadro constant. For $g$ equally populated internal species, exchange occurs within each species and the coefficient in terms of total density is smaller by a factor $g$.
\b[Why the two signs differ.] <Bosonic exchange symmetry> enhances the <probability> of nearby identical <bosons>, whereas <fermionic exchange symmetry> suppresses nearby identical <fermions> in the same internal state. In the dilute regime, the normalized equal-time <pair detection correlation> is
$$
g^{(2)}(r)=1+\eta e^{-2\pi r^2/\lambda^2}.
$$
The exchange part follows by squaring the Gaussian one-particle coherence factor $e^{-\pi r^2/\lambda^2}$. At zero separation this gives $g^{(2)}(0)=2$ for <bosons> and $0$ for one-component <fermions>. Integrating the exchange pair weight provides a direct coefficient check:
$$
B_2^{\mathrm{ex}}=-\frac12\int[g^{(2)}(r)-1]d^3r
=-\frac\eta2\left(\frac{\lambda^2}{2}\right)^{3/2}
=-\frac{\eta\lambda^3}{2^{5/2}}.
$$
Bunching therefore acts like an effective statistical attraction in the dilute <equation of state>, lowering the <pressure> needed at fixed density. Exclusion acts like an effective statistical repulsion, raising it. Neither description asserts an actual interparticle force. Different internal <fermion> species do not exclude each other from a common spatial location in the same way, explaining the importance of the species convention.
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